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fqq
12:20 AM
@AndrewMicallef maybe the formatting is a bit confusing, it just means "on the face where $x=0$, $\mathbf{K_1}=...$" etc
so it's the same as "$\mathbf{K}_1(0,y,z)=\dots$" etc
 
oh okay, that helps a bit :)
Although I'm also realising I'm well out of my depth with this maths, would you care to talk a bit more with me about some of the other symbols?
ie what the n hat is, and what nabla M(x) means?
 
 
2 hours later…
2:48 AM
@AndrewMicallef The n hat is the unit vector perpendicular to a given face and the nabla cross M is going to be the curl of the magnetization
 
2:59 AM
cheers @danielu, i guess nhat must be common notation for a surface normal then
so I guess khat, ihat jhat would then be the unit vector components of the face, khat being normal and i and j being the components parallel to the surface?
 
So ihat, jhat, and khat are the normal vectors to the face. nhat is just being used to represent the normal to a given face
When computing the surface current K, you do M cross nhat for each face, so whether nhat is ihat, jhat, khat, or their negatives depends on the face you're considering
The answer is considering the bar magnet to be a rectangle from (0, 0) to (a, b) (ignoring the height), so you have an x=0 face that goes from (0, 0) to (0, b), an x=a face that's (a, 0) to (a, b) and so on
 
ok, and you only need to deal with 4 of the faces because the other two faces cancel each other out. So you compute surface currents on x=0, x=a, y=0, y=b.
and the surface current is Magnitude of magnetisation * direction
* face normal
if say, I have a neodynium magnet with Br = 1.0 T, would that mean $M_0 = 1.0$?
I feel like maybe I should open up a new question, but I also feel like it'll get closed for being way too similar and the answer will be go learn more maths
 
3:21 AM
Well, the other two faces cancel out because the cross product k cross k is 0 because they're unit vectors in the same direction
I'm not sure about the magnetization value. I would think that the field and magnetization would be different values, but it's been quite a while since I've thought about magnetic materials
It may be worth noting that different materials could have different properties. I'm not sure if a neodymium magnet would follow along with the model of a uniform magnetization or not
 
Oh that is exactly what I needed!
 
It differs by a scaling of $\mu_0$, but you also add a factor back in when calculating the field, so you might be able to just sweep it under the rug
 
3:36 AM
yeah, and scales with the volume of the magnet, which makes intuitive sense
@danielunderwood probably worth noting, but I think for the moment I need to understand fundamentals before getting caught up in trying to worry about material properties :)
One, hopefully last thing, Does $\mathbf M(\mathbf x)$ denote a function of $\mathbf x$ that takes a vector as input and outputs a vector?
 
Yep, in this context, anything bolded like that will be a vector
 
awesome, I think I have enough to work the rest out on my own!
thank you so much
 
3:53 AM
I have lots of conceptual doubts related to kinetic theory
But i am asking here in pair of three
In 1-2 days interval
Is there any limit on asking questions?
 
123
4:52 AM
Hi All
 
5:15 AM
heya
 
 
3 hours later…
8:06 AM
@kaylimekay Hey! you were kinda right in the comments you were making
Did you see the answer?
Ah ... just saw your answer
*comment in the answer
 
0
Q: Improving a poor question by non-authors

GiorgioPThere are already many Q&A related to the improvement of a question by the original author. It is quite obvious and well documented how to proceed to improve the form of a question. However, I did not find guidelines or discussions about the case of poorly formulated questions that will probably ...

 
123
8:40 AM
I have question. In my book it said. Two Unlike parallel forces (P, -P) acting on a rigid, The moment is zero about point P/DE = Q/AD, where DE distance between - P and that point, point AD is distance P and that point.
My question is how unlike two parallel forces is different from couple of forces. But moment of couple of force is constant never zero. It looks same
Couple moment is constant, unlike parallel can be zero at point. But both situations looks same. Pls answer
Hello @JohnRennie sir
 
@MoreAnonymous Ah yes, no worries. Glad you got an answer that was helpful.
 
123
9:07 AM
I found that unlike parallel forces can be different in magnitude. What if unlike parallel forces are equal in magnitude, can we say them couple???
 
 
1 hour later…
123
10:07 AM
@JohnRennie Sir pls have a look my question when you have time.
 
10:20 AM
Hey, @JohnRennie your GitHub username on your profile is wrong!
 
10:43 AM
@bolbteppa I might be wrong but weren't you and ACuriousMind talking about the problem of time? Only to conclude there was no such problem?
 
10:56 AM
@KrrishDhaneja what
 
@KrrishDhaneja that was just the default value that the SE put in when I created the account. Anyhow, I've changed it to the correct URL though I don't have a lot of stuff on github. Thanks :-)
 
11:12 AM
@JohnRennie are you affiliated with Kodi?
 
@MoreAnonymous it was talked about in here before yeah, I wouldn't say we concluded anything but at least for now it's something I'm skeptical of
 
@NiharKarve no. I lost interest in it and stopped contributing so they took me off the team.
Which is fair enough.
 
I have just the post you should contribute to:

https://physics.stackexchange.com/questions/608225/why-is-there-a-problem-of-time-in-quantum-gravity
 
@MoreAnonymous the problem of time actually applies to any theory that is reparameterisation invariant, and not just GR.
But it's a surprisingly hard thing to explain in simple terms. I have actually attempted this before and ultimately deleted the answer because I felt it was usatisfactory.
If you have any system that is reparameterisation invariant then the Hamiltonian is always zero i.e. there can be no time evolution. This applies to classical systems as well as quantised ones.
 
The Wheeler–DeWitt equation is a field equation. It is part of a theory that attempts to combine mathematically the ideas of quantum mechanics and general relativity, a step towards a theory of quantum gravity. In this approach, time plays a role different from what it does in non-relativistic quantum mechanics, leading to the so-called 'problem of time'. More specifically, the equation describes the quantum version of the Hamiltonian constraint using metric variables. Its commutation relations with the diffeomorphism constraints generate the Bergman–Komar "group" (which is the diffeomorphism...
 
11:18 AM
This is a better link:
The Hamiltonian constraint arises from any theory that admits a Hamiltonian formulation and is reparametrisation-invariant. The Hamiltonian constraint of general relativity is an important non-trivial example. In the context of general relativity, the Hamiltonian constraint technically refers to a linear combination of spatial and time diffeomorphism constraints reflecting the reparametrizability of the theory under both spatial as well as time coordinates. However, most of the time the term Hamiltonian constraint is reserved for the constraint that generates time diffeomorphisms. == Simplest... ==
Although this applies to GR they give an example of a simple mechanical system where the same problem exists.
 
Yeah this is very confusing tbh
 
@bolbteppa yes, I fear that to attempt one of my trademark "explain it simply" answers would just reveal I don't understand it as well as I should.
 
legendary trademark, sir
 
:-)
 
@JohnRennie But Hamiltonians which are zero exist and studied as well (by witten)
In gauge theory and mathematical physics, a topological quantum field theory (or topological field theory or TQFT) is a quantum field theory which computes topological invariants. Although TQFTs were invented by physicists, they are also of mathematical interest, being related to, among other things, knot theory and the theory of four-manifolds in algebraic topology, and to the theory of moduli spaces in algebraic geometry. Donaldson, Jones, Witten, and Kontsevich have all won Fields Medals for mathematical work related to topological field theory. In condensed matter physics, topological quantum...
"The reason why a theory with a zero Hamiltonian can be sensibly formulated resides in the Feynman path integral approach to QFT. This incorporates relativistic invariance (which applies to general (d + 1)-dimensional "spacetimes") and the theory is formally defined by a suitable Lagrangian
—a functional of the classical fields of the theory. A Lagrangian which involves only first derivatives in time formally leads to a zero Hamiltonian, but the Lagrangian itself may have non-trivial features which relate to the topology of M."
 
11:35 AM
Yes, I'm just saying I doubt I could give a satisfactory answer to that question because the situation is more complicated than it appears at first sight.
 
P.S I am waaayyyyy to far from being an expert on these topics
 
"Deparametrization and the identification of a time variable with respect to which everything evolves is the opposite process of parametrization. It turns out in general that not all reparametrisation-invariant systems can be deparametrized. General relativity being a prime physical example... General relativity is an example of a physical theory where the Hamiltonian constraint isn't of the above mathematical form in general, and so cannot be deparametrized in general."
From the Hamiltonian Constraint link, this is why it may be a serious issue I'm just not sure
 
11:49 AM
Mornin everyone
 
 
2 hours later…
1:48 PM
a certain medium with permittivity equal to that of vacuum occupies the space between two conducting slabs located at y = ±2 cm. when heated, the material emits electrons such that their charge density is given by ρ = 50(1−y^2 ) μc/m3 . if both the slabs are held at 30 kv, find the potential distribution within the slabs.
using poisson's equation , if we assume V to be a function of y only, we can solve for V(y) eaily, with 2 arbitrary constants, that can be found out using v(2)=v(-2)=30
However, how do I rigorously justify that V is not a function of x or ?
 
2:03 PM
@satan29 hi
 
@JackRod hi
btw, I think I got : its simply the uniqueness theorem isnt it?
If i find A solution, it is The solution. right?
 
 
2 hours later…
fqq
3:44 PM
@satan29 yes, symmetry + uniqueness
 
123
4:25 PM
When book says moment about a point. Does it mean it is always a fixed point or not??? Pls clear
 
4:43 PM
A couple of things that might be of interest. Weinberg has a new paper out On the Development of Effective Field Theory.
And on the same topic the book Introduction to Effective Field Theory has been made free online.
 
5:07 PM
Also, for those of us old enough to remember John Baez's ur-blog apparently he is getting them properly typeset and published as pdfs on the Arxiv. The first batch is here: This Week's Finds in Mathematical Physics (1-50).
Though part of the charm was that all his diagrams were ASCII art. He even used to draw Feynman diagrams in ASCII.
 
Hell yeah
I'm a huge John Baez fan
 
vzn
baez is great/ legendary. he can still put his ASCII art in the paper(s), lets see :)
poking around now again. what does everyone else like? hes deep into category theory. looking for something interesting. "finds in the finds?" how about bitcoin 2014! post by adleman? :) johncarlosbaez.wordpress.com/2014/01/27/…
 
vzn
5:32 PM
ah way cool pythagorean triples number theory problem by Heule on SAT was just reading about this :) johncarlosbaez.wordpress.com/2020/04/13/bigness-part-2 phys.org/news/2016-05-math-proof-largest-terabytes.html cs.utexas.edu/~marijn/ptn
 
 
2 hours later…
7:28 PM
Hey all, any meteorologists around? Or someone with an undergrad or above understanding of meteorology?
 
I only sacrifice animals to please the sky gods for meteorology
 
7:42 PM
:(
 
8:33 PM
Hope this is a reasonable question, but when we derive the EL equations in field theory and discard the boundary term in our integration by parts we assume the field vanishes at spatial infinity. But, for instance, the simplest classical field we use, the Klein-Gordon field, has solutions that are plane waves, which don't decay at infinity, they just oscillate forever, so why are we justified in making this assumption?
 
@Charlie Plane wave solutions aren't usually assumed to be physical
We use them because they form a basis of the space of solutions
 
Oh yeah I guess that's fair
so our physical solutions are more like blips in the field centred around a point, rathe than a "sea" of oscillations
 
ie plane waves are a basis of $L^2$ functions
Yes
usually we assume that functions are typically something like Schwarz functions
 
I haven't gone out of my way to learn about function spaces and their linear structure, seems like it's fairly invaluable in places
 
that decay to 0 at infinity
 
8:35 PM
ahhh
 
of course, you can do field theory with fields that do not decay at infinity, but then things are more complicated
 
Would that be a fairly niche sector of physics or am I unlikely to encounter than in, for instance qft
 
The common example is general relativity
 
oh, lol
 
It's common in GR to assume that the field may not vanish at infinity
In this case, we remove the boundary term by adding an extra term to the Lagrangian
 
8:37 PM
oh is that in things like de-Sitter space with constant curvature
 
The so called Hawking-Gibbon term
For instance, yes
In general relativity, the Gibbons–Hawking–York boundary term is a term that needs to be added to the Einstein–Hilbert action when the underlying spacetime manifold has a boundary. The Einstein–Hilbert action is the basis for the most elementary variational principle from which the field equations of general relativity can be defined. However, the use of the Einstein–Hilbert action is appropriate only when the underlying spacetime manifold M {\displaystyle {\mathcal {M}}} is closed, i.e., a manifold which is...
 
hey guys, is there a nice way to think about faraday cages based on antenna EM diagrams
 
I haven't sunk much time into GR in a while, getting rusty already
I see, ty
 
You can write the GR Lagrangian in a more compact way with forms, IIRC
So that you don't have to put in the weird surface integral
 
8:55 PM
@Slereah is that from some gauss divergence thm shenanigans
 
Pretty much, yeah
Stokes theorem, in general
 
theres the name i was looking for
 
 
3 hours later…
11:38 PM
hi all, I would appreciate some feedback on this answer: physics.stackexchange.com/a/608256/84967
4
the first sentence is very much false (and it does not answer the issue in the OP)
and the rest is just a list of (irrelevant) references
should it be flagged as not-an-answer?
I fear reviewers will most likely miss the context and vote as looks-fine
Thanks!
 

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